POJ3295--Tautology

转自：http://www.cnblogs.com/lv-2012/archive/2012/10/27/2742751.html

Tautology

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7220 Accepted: 2741

Description

WFF 'N PROOF is a logic game played with dice. Each die has six faces representing some subset of the possible symbols K, A, N, C, E, p, q, r, s, t. A Well-formed formula (WFF) is any string of these symbols obeying the following rules:

• p, q, r, s, and t are WFFs
• if w is a WFF, Nw is a WFF
• if w and x are WFFs, Kwx, Awx, Cwx, and Ewx are WFFs.

The meaning of a WFF is defined as follows:

• p, q, r, s, and t are logical variables that may take on the value 0 (false) or 1 (true).
• K, A, N, C, E mean and, or, not, implies, and equals as defined in the truth table below.

Definitions of K, A, N, C, and E     w  x  Kwx  Awx   Nw  Cwx  Ewx  1  1  1  1   0  1  1  1  0  0  1   0  0  0  0  1  0  1   1  1  0  0  0  0  0   1  1  1

A tautology is a WFF that has value 1 (true) regardless of the values of its variables. For example, ApNp is a tautology because it is true regardless of the value ofp. On the other hand, ApNq is not, because it has the value 0 for p=0, q=1.

You must determine whether or not a WFF is a tautology.

Input

Input consists of several test cases. Each test case is a single line containing a WFF with no more than 100 symbols. A line containing 0 follows the last case.

Output

For each test case, output a line containing tautology or not as appropriate.

Sample Input

ApNpApNq0

#include<iostream>#include<string>#include<stack>using namespace std;string str;stack<int> stk;bool v[6];bool tautology;bool getSult(){int a,b;for(int i=str.length()-1;i>=0;i--){switch(str[i]){case 'K':a=stk.top();stk.pop();b=stk.top();stk.pop();stk.push(a&&b);break;case 'A':a=stk.top();stk.pop();b=stk.top();stk.pop();stk.push(a||b);break;case 'N':a=stk.top();stk.pop();stk.push(!a);break;case 'C':a=stk.top();stk.pop();b=stk.top();stk.pop();stk.push(!a||b);break;case 'E':a=stk.top();stk.pop();b=stk.top();stk.pop();stk.push(a==b);break;default:stk.push(v[str[i]-'p']);}}return stk.top();}int main(){int i,j;while(cin>>str&&str!="0"){while(!stk.empty())stk.pop();tautology=true;for(i=0;i<31;i++){for(j=0;j<5;j++)v[j]=(i>>j)%2;if(!getSult()){tautology=false;break;}}if(tautology){cout<<"tautology"<<endl;}elsecout<<"not"<<endl;}return 0;}